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CAT 2025 Lesson : Progressions - Concepts & Cheatsheet

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Note: The video for this module contains a summary of all the concepts covered in this lesson. The video would serve as a good revision. Please watch this video in intervals of a few weeks so that you do not forget the concepts. Below is a cheatsheet that includes all the formulae but not necessarily the concepts covered in the video.

  7. Cheatsheet

11) Where a\bm{ a } and d\bm{ d } are the first term and common difference respectively in an AP with n\bm{n} terms,

(a)nth\bm{n^{th}} term = aa + (nn – 1)dd

(b) Average of an AP = Average of First and Last terms = x1+xn2\dfrac {x_1 + x_n}{2}

(c) Sum of terms of an AP = Sn\bm{ S_n} = n×Averagen \times Average

= n×(x1+xn2)=n2×[2a+(n−1)d]n \times \left( \dfrac{x_1 + x_n}{2} \right) = \dfrac {n}{2} \times [2a + (n - 1)d]

(d) Number of terms in an AP = n\bm{ n }

= Last Term−First TermCommon Difference+1\dfrac {Last \ Term - First \ Term}{Common \ Difference} + 1

= xn−x1d+1\dfrac {x_n - x_1}{d} + 1

22) Where a\bm{a} and r\bm{ r} are the first term and common ratio respectively in a GP with n terms,

(a) nth\bm{ n^{th}} term = arn−1ar^{n-1}

(b) Geometric Mean of GP = GM of First and Last terms =x1×xn\sqrt{x_1 \times x_n}

= a×ar(n−1)\sqrt{a \times ar^{(n-1)}} = ar(n−1)/2ar^{(n - 1)/2}

(c) Sum of n terms of a GP = Sn\bm{ S_n } = a(rn−1)r−1\dfrac{a(r^{n}-1)}{r-1}

(d) If a GP has infinite terms and 0<r<10 \lt r \lt 1, then Sum of infinite terms of the GP = a1−r\dfrac{a}{1 - r}

33) A sequence x1,x2,x3,....,xnx_1,x_2,x_3,....,x_n is said to be in Harmonic Progression if 1x1,1x2,1x3,.....,1xn\dfrac{1}{x_1},\dfrac{1}{x_2},\dfrac{1}{x_3},.....,\dfrac{1}{x_n} are in Arithmetic Progression.


(a) If a,b,c a, b, c are in HP, then 1b−1a=1c−1b\dfrac{1}{b} - \dfrac{1}{a} = \dfrac{1}{c} - \dfrac{1}{b} and b=2aca+cb = \dfrac{2ac}{a + c}

(b) In an HP, the middle term is the Harmonic Mean.

44) Sum of first n\bm{ n } natural numbers = n(n+1)2\dfrac{n(n + 1)}{2}

55) Sum of squares of first nn natural numbers = (n(n+1)(2n+1))6\dfrac{(n(n + 1)(2n + 1))}{6}

66) Sum of cubes of first n\bm{n} natural numbers = [n(n+1)2]2\left[ \dfrac{n(n + 1)}{2} \right]^2

77) Sum of first n \bm{n } even numbers =n(n+1)n (n + 1)

88) Sum of first n\bm{n } odd numbers = n2n^2
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